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Precipitation Reactions - solubility table_ net ionic equations

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330 04’02: PRECIPITATION REACTIONS

CORRELATIONS
National Content Standards:
Connecticut Standards:
Ridgefield Themes:


ESSENTIAL CONCEPT
“In a precipitation reaction, the product, an insoluble substance, separates from the solution. Acid-base
reactions involve the transfer of a proton (H+) from an acid to a base. In an oxidation-reduction reaction,
or redox reaction, electrons are transferred from a reducing agent to an oxidizing agent. These three types
of chemical reactions represent the majority of reactions in chemical and biological sytems.”1

HOMEWORK
DUE:
DO:                                   4.19 – 4.24


REVIEW:
1.     Into
2.     Atoms, Molecules, and Ions (elements = letters; formula = word; reaction = sentence)
3.     Stoichiometry (grammar)
4.     Reactions in Aqueous Solutions (starting to write a story)
       4.1 electrolyte/nonelectrolyte; dissociation/ionization; solute/solvent/solution/alloy; hydration;
       chemical equilibria/reversible reaction

LESSON:




1
     Yang, R. (2008). Chapter 3: Reactions in Aqueous Solutions. In General Chemistry: The Essential Concepts (5th
     ed., p. 94). New York: McGraw Hill.
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§04’02: PRECIPITATION REACTIONS

1. OVERVIEW
   A. One way to classify chemical reactions (Table 1):
  Table 1. Five General Types of Chemical Reactions
             Type of Reaction                General Equation                                      Example
    synthesis                               A + B  AB                                   2H2 + O2  2H2O
    decomposition                             AB  A + B                                 2HgO  2Hg + O2
    single replacement
         cation single replacement        A + BX  AX + B                2Li + 2H2O  2LiOH + H2
         anion single replacement         A + YB  YA + B                F2 + 2NaBr 2NaF + Br2
    double replacement (metathesis)     AX + BY  AY + BX                KCN + HBr  KBr + HCN
      combustion                                   CxHy + O2  CO2 + H2O CH4 + 2O2  CO2 + 2H2O
  In the first four types of reactions, A, B, C, D, X, and Y represent elements or polyatomic ions. Hence, the formation of water
  from hydrogen and oxygen gas (2H2 + O2  2H2O) is a synthesis reaction and fits the general format, A+B  AB.


    B. One common type of reaction is precipitation reaction
       i.  results in the formation of an insoluble product (a.k.a., precipitate)
       ii. example: Pb(NO3)2(aq) + 2KI(aq)  PbI2(s) + 2KNO3(aq)
           (which, in this case, is a double replacement reaction)
                                                                                               o
  Table 2. Solubilities Rules for Common Ionic Compounds in Water (25 C)
   Compounds Containing:
                  Soluble Compounds                                 Exceptions
                        +   +  +    +    +
    alkali metals (Li , Na , K , Rb , Cs )
                         +
    ammonium (NH4 )
                   -
    nitrates (NO3 )
                                            -
    hydrogen carbonates, bicarbonates (HCO3 )
                      -
    chlorates (ClO3 )
                -     - -                                      +     2+    2+
    halides (Cl , Br , I )                      halides of Ag , Hg2 , Pb
                   2-
    sulfates (SO4 )                             sulfates of Ag+, Hg22+, Pb2+, Ca2+, Sr2+, Ba2+

                 Insoluble Compounds                                        Exceptions
     carbonates (CO32-)
                         2-
     chromates (CrO4 )
                 2-                                     alkali metals ions, ammonium ion
     sulfides (S )
                        3-
     phosphates (PO4 )
                       -
     hydroxides (OH )                                  alkali metal ions, ammonium, Ba2+
    useful help: soluble: all alkali metals, ammonium, nitrates, bicarbonates, chlorate

            Example 4.1
            Classify the following ionic compounds as soluble or insoluble:
            a. silver sulfate (Ag2SO4)
            b. calcium carbonate (CaCO3)
            c. sodium phosphate (Na3PO4)
            [answers]
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2. Solubility
   A.  the maximum amount of solute that will dissolve in a given quantity of solvent at a specific
       temperature

    B. Influence of temperature and state of matter:
       i. solids in aqueous solutions: solubility increases with increasing temperature
                                        (e.g., dissolving sugar in hot coffee vs. cold iced tea)
       ii. gases in aqueous solutions: solubility decreases with increasing temperatures
                                       (e.g., soda from refrigerator vs. in sun at beach in summer)

3. Molecular Equations, Ionic Equations, and Net Ionic Equations
   A. Molecular Equations = formulas of compounds are written as though all species existed as
                             molecules or whole uints. (what we’ve seen so far)

    B. Ionic compounds dissolve in water to produce their component species:
       molecular equation: Pb(NO3)2(aq) + 2KI(aq)  PbI2(s) + 2KNO3(aq)

                                                   + 2NO3–(aq) + 2K         2I–(aq)
                                          2+                        +
         ionic equation:               Pb   (aq)                        (aq) +
                                                                            PbI2(s) + 2K (aq) + NO3–(aq)
                                                                                         +


    C. net ionic equation: Because some ions (spectator ions) appear on both sides of the equation, we
       can cancel them to produce the net ionic equation which shows only the species that actually
       take part in the reaction:
                                  Pb (aq) + 2NO3–(aq) + 2K (aq) + 2I–(aq)
                                    2+                    +
       ionic equation:
                                                                                                     + NO3–(aq)
                                                                                          +
                                                                            PbI2(s) + 2K     (aq)


                                                   + 2I–(aq)  PbI2(s)
                                          2+
         net ionic equation:           Pb   (aq)


         i.   Steps for writing ionic and net ionic equations:
                  (i). Write a balanced molecular equation (correct formulas & balanced). Determine
                         which, if any, precipitates will be produced
                  (ii). Write the ionic equation. The compound that doesn’t form a precipitate is written as
                         free ions.
                  (iii). Cancel spectator ions. Write the net ionic equation for the reaction.
                  (iv). Check: charges, and number of atoms
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Example 4.2 (modified)
Predict the reaction between potassium phosphate and calcium nitrate in water.

A. Set up the problem
        1. Formulas:
                                                   K3PO4
                                       +     3–
                potassium phosphate = K + PO4
                                                –
                                   = Ca + NO3  Ca(NO3)2
                                        2+
                calcium nitrate

        2. Predict reaction:
                a. From the beginning of the reaction (K3PO4 + Ca(NO3)2 ), we see that the reaction is
                   a double displacement reaction and we can write the unbalanced chemical equation:
                2K3PO4(aq) + 3Ca(NO3)2(aq)  1Ca3(PO4)2(aq) + 6KNO3(aq)

                  b. The two reactants are soluble in water (see above), so they dissociate into their ionic
                     species:
                          K3PO4  3K+ + 1PO43–
                          Ca(NO3)2  1Ca2+ + 2NO3–

Now we’re set to solve the problem:
1. Write the balanced molecular equation:
      2K3PO4(aq) + 3Ca(NO3)2(aq)  1Ca3(PO4)2(aq) + 6KNO3(aq)

2. Write the ionic equation. (Soluble compounds dissociate into ions.)
                                           –                         –
      6K (aq) + 2PO4 (aq) + 3Ca + 6NO3 (aq)  6K (aq) + 6NO3 (aq) + Ca3(PO4)3()
          +            3–          2+                 +


3. Cancel out spectator ions to obtain net ionic equation:
      3Ca2+(aq) + 2PO43–(aq)  1Ca3(PO4)3()

4. Check:
   A. net ionic equation is balanced
   B. charges balance (positives {Ca: 3*+2 = PO4: 2*-3)

Practice:
Predict the reaction of aluminum nitrate + sodium hydroxide. {answer}
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Example 4.1
Classify the following ionic compounds as soluble or insoluble:
a. silver sulfate (Ag2SO4)              .... sulfates are generally soluble, except sulfates of Ag+, Hg22+,
                                       Pb2+, Ca2+, Sr2+, Ba2+ - so silver(I) sulfate is insoluble
b. calcium carbonate (CaCO3)            .. carbonates are generally insoluble except alkali metals & NH4+
                                                 .................................... this is none of those so it is insoluble
c. sodium phosphate (Na3PO4)                     ................................... alkali metal ion compounds are all soluble


Example 4.2 Practice Problem:
Aluminum nitrate = Al(NO3)3; sodium hydroxide = NaOH; double replacement reaction

1. Balanced Chemical Equation:
              1Al(NO3)3(aq) + 3NaOH(aq)  1Al(OH)3(s) + 3NaNO3(aq)

2. Ionic Equation:
               1Al3+(aq) + 3NO3–(aq) + 3Na+(aq) + 3OH– (aq)  1Al(OH)3(s) + 3Na+(aq) + 3NO3–(aq)

3. Net Ionic Reaction:
               1Al3+(aq) + 3OH– (aq)  1Al(OH)3(s)

4. Check:          number of atoms {3Al = 3Al}
                   charges {1 * (+3) = +3; 3 * {-1} = -3}

				
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